Find solution to system with logarithms I have two equations:
\begin{align*}
3 \ln x + \ln y &= 3 \\
4 \ln x - 6 \ln y &= -7 \\
\end{align*}
Do I just proceed as I have learned with adding equations resulting in:
\begin{align*}
18 \ln x + 6 \ln y &= 18 \\
4 \ln x - 6 \ln y &= -7 \\
22 \ln x &= 11 \\
\end{align*}
Then solve from there?
 A: Yes, that is how the system is handled. In a more complicated situation, we might want to make a substitution, letting $u=\ln x$ and $v=\ln y$. Then our two equations become
$$3u +v=3, \qquad 4u-6v=-7.$$
This is a system of two linear equations in the two unknowns $u$ and $v$.
If we have a system of quite a few linear equations in quite a few unknowns, we need to adopt a systematic approach.  For our simple system, almost anything sensible will work. For example, we can try to "eliminate" $v$ (that was your strategy). Multiply each side of the first equation by $6$. We obtain $18u+3v=18$.  
From the equations $18u+3v=18$ and $4u-6v=-7$, we obtain, by adding, that $22u=11$, so $u=1/2$.  Substituting this in one of the equations, we get $v=3/2$.
So $\ln x=1/2$ and $\ln y=3/2$. It follows that $x=e^{1/2}$ and $y=e^{3/2}$.
Comment: As noted above, for this simple system almost any reasonable strategy will work. For example, in high school one might be encouraged to rewrite the first equation as $v=3-3u$, and then substitute for $v$ in the second equation. We would get
$$4u-6(3-3u)=-7,$$ 
which, after some manipulation, yields $22u=11$. 
